Find the values of x if p + 7 = 0, q - 12 = 0 and x^2 + px + q = 0. If x = p is a solution of the equation x(2x + 5) = 3, then find the values of p. If x = 3 is a solution of the e... Find the values of x if p + 7 = 0, q - 12 = 0 and x^2 + px + q = 0. If x = p is a solution of the equation x(2x + 5) = 3, then find the values of p. If x = 3 is a solution of the equation (k + 2)x^2 - kx + 6 = 0, find the value of k. Hence, find the other root of the equation. Read more

Steps to Solve

1. Solve the first set of equations for ( p ) and ( q )

The equations given are ( p + 7 = 0 ) and ( q - 12 = 0 ).

To find ( p ) and ( q ):

• From ( p + 7 = 0 ): $$ p = -7 $$

• From ( q - 12 = 0 ): $$ q = 12 $$

2. Substitute ( p ) into the equation ( x(2x + 5) = 3 )

We substitute ( p = -7 ) into the equation: $$ x(2x + 5) = 3 $$

Now expand and rearrange the equation: $$ 2x^2 + 5x - 3 = 0 $$

3. Use the quadratic formula to find ( x )

The quadratic formula is given by: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

For the equation ( 2x^2 + 5x - 3 = 0 ):

• Coefficients are: ( a = 2 ), ( b = 5 ), ( c = -3 )

Plugging these into the quadratic formula: $$ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} $$

Calculate the discriminant: $$ 5^2 - 4 \cdot 2 \cdot (-3) = 25 + 24 = 49 $$

So we have: $$ x = \frac{-5 \pm \sqrt{49}}{4} $$ $$ x = \frac{-5 \pm 7}{4} $$

Thus, the solutions are:

1. $$ x = \frac{2}{4} = \frac{1}{2} $$

2. $$ x = \frac{-12}{4} = -3 $$

3. Find the value of ( k ) using ( x = 3 ) in ( (k + 2)x^2 - kx + 6 = 0 )

Substituting ( x = 3 ) into ( (k + 2)(3^2) - k(3) + 6 = 0 ): $$ (k + 2)(9) - 3k + 6 = 0 $$ Expanding this gives: $$ 9k + 18 - 3k + 6 = 0 $$

Combining like terms: $$ 6k + 24 = 0 $$

Now solving for ( k ): $$ 6k = -24 $$ $$ k = -4 $$

5. Calculate the other root of the quadratic equation

To find the other root, use the sum of the roots obtained from ( (k + 2)x^2 - kx + 6 = 0 ) with ( k = -4 ): $$ (k + 2) \text{ at } k = -4 $$ becomes ( -2x^2 + 4x + 6 = 0 ).

Using the sum of the roots:

• Let the roots be ( r_1 ) and ( r_2 ). We know ( r_1 + r_2 = -\frac{b}{a} = -\frac{4}{-2} = 2 ). Since ( r_1 = 3 ), we have: $$ 3 + r_2 = 2 $$ $$ r_2 = 2 - 3 = -1 $$